Lecture 06

Bill Perry

Lecture 6: Review

Covered

Introduction to hypothesis testing
The standard normal distribution
Standard error
Confidence intervals
Student’s t-distribution
H testing
One and Two Sample T Test
p-values

Lecture 6: Overview

The objectives:

p-values
Brief review
H test for a single population
1- and 2-sided tests
Hypothesis tests for two populations
Assumptions of parametric tests

Lecture 6: Statistical hypothesis testing

Major goal of statistics:
- inferences about populations from samples…
  - assign degree of confidence to inferences
- Statistical hypothesis testing:
  - formalized approach to inference
- Hypotheses ask whether samples come from populations with certain properties
- Often interested in questions about population means
  - but other questions are of interest

Lecture 6: Statistical hypothesis testing

Useful hypotheses: - Rely on specifying - null hypothesis (Ho) - alternate hypothesis (Ha)

Ho is the hypothesis of “no effect”
- two samples from population with same mean
- sample is from population of mean = 0
Ha (research hypothesis)
- is the opposite of the Ho
- or predicts that there is an effect of x on y
- but does NOT suggest a direction

Lecture 6: Statistical hypothesis testing

Together Ho and Ha encompass all possible outcomes:

For Example:
Ho: µ=0, Ha: µ ≠ 0
- mean equals 0 or mean does not equal 0
Ho: µ=35, Ha: µ ≠ 35
- mean equals 35 or mean does not equal 35
- Ho: µ1 = µ2, Ha: µ1 ≠ µ2
- mean of population 1 equals mean of population 2 or it does not
- Ho: µ > 0, Ha: µ ≤ 0
- can be directional mean is greater than 0 or mean is not equal or less than 0

Lecture 6: Statistical hypothesis testing

Tests assess likelihood of the null hypothesis being true

If the Ho is likely false, then Ha assumed to be correct
More precisely:
- the long run probability of obtaining sample value (or more extreme one) if the null hypothesis is true
  - p(data|Ho) - the probability of observing the data given that the null hypothesis Ho is true

Lecture 6: Statistical hypothesis testing

Hypothesis tests

Expressed as p-value (0 to 1)
Interpret p-value as:
- probability of obtaining sample value of statistic (or more extreme one) if Ho is true
High p-value:
- high probability of obtaining sample statistic under Ho
  - if the null hypothesis (Ho) were true, you would frequently observe data similar to or more extreme than your sample statistic
  - your observed results are quite compatible with what the null hypothesis predicts
- low p-value: low probability of obtaining sample statistic under Ho
  - if the null hypothesis (Ho) were true, you would rarely observe data similar to or more extreme than your sample statistic
  - Your results are unusual under the null hypothesis, suggesting that either you’ve witnessed a rare event or the null hypothesis may be incorrect

Lecture 6: Statistical hypothesis testing

Statistical test results:

p = 0.3 means that if I repeated the study 100 times, I would get this (or more extreme) result due to chance 30 times

p = 0.03 means that if I repeated the study 100 times, I would get this (or more extreme) result due to chance 3 times

Which p-value suggests Ho likely false?

Lecture 6: Statistical hypothesis testing

Statistical test results:

At what point reject Ho?

p < 0.05 conventional “significance threshold” (α = alpha or p value)

p < 0.05 means: if Ho is true and we repeated the study 100 times - we would get this (or more extreme) result less than 5 times due to chance

Lecture 6: Statistical hypothesis testing

Statistical test results:

α is the rate at which we will reject a true null hypothesis (Type I error rate)

Lowering α will lower likelihood of incorrectly rejecting a true null hypothesis (e.g., 0.01, 0.001)

Both Hs and α are specified BEFORE collection of data and analysis

Lecture 6: Statistical hypothesis testing

Traditionally α=0.05 is used as a cut off for rejecting null hypothesis

There is nothing magical about 0.05 - actual p-values need to be reported - also need to decide prior to study

p-value range	Interpretation
P > 0.10	No evidence against Ho - data appear consistent with Ho
0.05 < P < 0.10	Weak evidence against the Ho in favor of Ha
0.01 < P < 0.05	Moderate evidence against Ho in favor of Ha
0.001 < P < 0.01	Strong evidence against Ho in favor of Ha
P < 0.001	Very strong evidence against Ho in favor of Ha

Lecture 6: Statistical hypothesis testing

Fisher:

p-value as informal measure of discrepancy between data and Ho

“If p is between 0.1 and 0.9 there is certainly no reason to suspect the hypothesis tested. If it is below 0.02 it is strongly indicated that the hypothesis fails to account for the whole of the facts. We shall not often be astray if we draw a conventional line at .05 …”

Lecture 6: Statistical hypothesis testing

General procedure for H testing:

Specify Null (Ho) and alternate (Ha)
Determine test (and test statistic) to be used
Test statistic is used to compare your data to expectation under Ho (null hypothesis)
Specify significance (α or p value) level below which Ho will be rejected

Lecture 6: Statistical hypothesis testing

General procedure for H testing:
Collect data
Perform test
- If p-value < α, conclude Ho is likely false and reject it
- If p-value > α, conclude no evidence Ho is false and retain it

Lecture 6: Brief review

Recall…

Major goal of statistics: inferences about populations from samples… and assign degree of confidence to inferences
Statistical H-testing: formalized approach to inference
Relies on specifying null hypothesis (Ho) and alternate hypothesis (Ha
Tests assess likelihood of the null hypothesis being true
Expressed as p-value: probability of obtaining sample value of statistic (or more extreme one) if Ho is true

Lecture 6: Brief review

Recall pine needle example

Probability of getting sample
with ȳ at least as far away from 21 as 35)? - p(ȳ ≤ 3500 or ȳ ≥ 3900)
- What about - 1-tailed or 2-tailed test?
- Can solve using SND and z-scores

Lecture 6: Brief review

z= (21-35)/40 = -0.48
- From z table: p= 0.6368 X 2
- p of getting sample as far away from µ as A is = 0.6368 (63.6%)
But - usually can’t use z!
Can use t-distribution instead…

Pine Needle Length: Hypothesis Testing Activity

This activity will guide you through the process of conducting single-sample and two-sample t-tests on pine needle data. We’ll explore how environmental factors like wind exposure might affect pine needle length.

You’ll learn to:

Formulate hypotheses
Test assumptions
Perform t-tests
Visualize data
Report results accurately

Part 1: Single Sample T-test

A single sample t-test asks whether a population parameter (like \(\bar{x}\)) differs from some expected value.

The question: Is the average pine needle length from our windward sample different from 55mm?

One-sample t-test

Used when we want to compare a sample mean to a known or hypothesized population value.

\(t = \frac{\bar{x} - \mu}{s/\sqrt{n}}\)

where:

\(\bar{x}\) is the sample mean
\(\mu\) is the hypothesized population mean
\(s\) is the sample standard deviation
\(n\) is the sample size

How to do this in R

# Install packages if needed (uncomment if necessary)
# install.packages("readr")
# install.packages("tidyverse")
# install.packages("car")
# install.packages("here")

# Load libraries
library(car)          # For diagnostic tests
library(tidyverse)    # For data manipulation and visualization

# Load the pine needle data
# Use here() function to specify the path
pine_data <- read_csv("data/pine_needles.csv")

# Examine the first few rows
head(pine_data)

# A tibble: 6 × 6
  date    group       n_s   wind  tree_no len_mm
  <chr>   <chr>       <chr> <chr>   <dbl>  <dbl>
1 3/20/25 cephalopods n     lee         1     20
2 3/20/25 cephalopods n     lee         1     21
3 3/20/25 cephalopods n     lee         1     23
4 3/20/25 cephalopods n     lee         1     25
5 3/20/25 cephalopods n     lee         1     21
6 3/20/25 cephalopods n     lee         1     16

grayling_df <- read_csv("data/gray_I3_I8.csv")

Part 1: Exploratory Data Analysis

Before conducting hypothesis tests, we should always explore our data to understand its characteristics.

Let’s calculate summary statistics and create visualizations.

Activity: Calculate basic summary statistics for pine needle length

# YOUR TASK: Calculate summary statistics for pine needle length
# Hint: Use summarize() function to calculate mean, sd, n, etc.

# Create a summary table for all pine needles
pine_summary <- pine_data %>%
  summarize(
    mean_length = mean(len_mm),
    sd_length = sd(len_mm),
    n = n(),
    se_length = sd_length / sqrt(n)
  )

print(pine_summary)

# A tibble: 1 × 4
  mean_length sd_length     n se_length
        <dbl>     <dbl> <int>     <dbl>
1        17.7      3.53    48     0.509

# Now calculate summary statistics by wind exposure
# YOUR CODE HERE

Part 1: Visualizing the Data

Activity: Create visualizations of pine needle length

Create a histogram and a boxplot to visualize the distribution of pine needle length values.

Effective data visualization helps us understand:

The central tendency
The spread of the data
Potential outliers
Shape of distribution

Your Task

# YOUR TASK: Create a histogram of pine needle length
# Hint: Use ggplot() and geom_histogram()

# Histogram of all pine needle lengths
ggplot(pine_data, aes(x = len_mm)) +
  geom_histogram(binwidth = 2, fill = "steelblue", color = "black") +
  labs(title = "Distribution of Pine Needle Length",
       x = "Length (mm)",
       y = "Frequency") +
  theme_minimal()

# Boxplot of pine needle length by wind exposure
# YOUR CODE HERE

Part 1: Single Sample T-Test

We want to test if the mean pine needle length on the windward side differs from 55mm.

Activity: Define hypotheses and identify assumptions

H₀: μ = 55 (The mean pine needle length on windward side is 55mm) H₁: μ ≠ 55 (The mean pine needle length on windward side is not 55mm)

Assumptions for t-test:

Data is normally distributed
Observations are independent
No significant outliers

Part 1: Testing Assumptions

Before conducting our t-test, we need to verify that our data meets the necessary assumptions.

Activity: Test the normality assumption

Methods to test normality:

Visual methods:
- QQ plots or histograms
- Statistical tests: Shapiro
- Wilk test

Assumptions in R - qqplots

# Filter for just windward side needles
windward_data <- pine_data %>% 
  filter(wind == "wind")

# YOUR TASK: Test normality of windward pine needle lengths
# QQ Plot
qqPlot(windward_data$len_mm, 
       main = "QQ Plot for Windward Pine Needles",
       ylab = "Sample Quantiles")

[1] 21 22

Shapiro Wilk

# Shapiro-Wilk test
shapiro_test <- shapiro.test(windward_data$len_mm)
print(shapiro_test)


    Shapiro-Wilk normality test

data:  windward_data$len_mm
W = 0.96062, p-value = 0.451

# Check for outliers using boxplot
# YOUR CODE HERE

Part 1: Conducting the Single Sample T-Test

Now that we’ve checked our assumptions, we can perform the single sample t-test.

Activity: Conduct a single sample t-test to compare windward needle length to 55mm What is probability of getting sample at least as far from 55mm as our sample mean?

This is our p-value, which helps us decide whether to reject the null hypothesis.

# Calculate summary statistics for windward needles
windward_summary <- windward_data %>%
  summarize(
    mean_length = mean(len_mm),
    sd_length = sd(len_mm),
    n = n(),
    se_length = sd_length / sqrt(n)
  )

print(windward_summary)

# A tibble: 1 × 4
  mean_length sd_length     n se_length
        <dbl>     <dbl> <int>     <dbl>
1        14.9      1.91    24     0.390

Your Task

# YOUR TASK: Conduct a single sample t-test
t_test_result <- t.test(windward_data$len_mm, mu = 55, var.equal = TRUE )
print(t_test_result)


    One Sample t-test

data:  windward_data$len_mm
t = -102.85, df = 23, p-value < 2.2e-16
alternative hypothesis: true mean is not equal to 55
95 percent confidence interval:
 14.11050 15.72284
sample estimates:
mean of x 
 14.91667

# Calculate t-statistic manually 
# YOUR CODE HERE: t = (sample_mean - hypothesized_mean) / (sample_sd / sqrt(n))

# can you do this manually or manually with R?

Part 1: Interpreting and Reporting Results

Activity: Interpret the t-test results

What does the p-value tell us?
Should we reject or fail to reject the null hypothesis?

How to report this result in a scientific paper:

“A two-tailed, one-sample t-test at α=0.05 showed that the mean pine needle length on the windward side (… mm, SD = …) [was/was not] significantly different from the expected 55 mm, t(…) = …, p = …”

Part 2: Two Sample T-Test

Now, let’s compare pine needle lengths between windward and leeward sides of trees.

Question: Is there a significant difference in needle length between the windward and leeward sides?

This requires a two-sample t-test.

Two-sample t-test compares means from two independent groups.

\(t = \frac{\bar{x}_1 - \bar{x}_2}{S_p\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}\)

where:

x̄₁ and x̄₂: These represent the sample means of the two groups you’re comparing
s²ₚ: This is the pooled variance, calculated as: s²ₚ = [(n₁ - 1)s₁² + (n₂ - 1)s₂²] / (n₁ + n₂ - 2), where s₁² and s₂² are the sample variances of the two groups.
n₁ and n₂: These are the sample sizes of the two groups.
√(1/n₁ + 1/n₂): This represents the pooled standard error.

Part 2: Exploratory Data Analysis by Group

Activity: Calculate summary statistics grouped by wind exposure Before conducting the test, we need to understand the data for each group.

# YOUR TASK: Calculate summary statistics by wind exposure
# Hint: Use group_by() and summarize()

group_summary <- pine_data %>%
  group_by(wind) %>%
  summarize(
    mean_length = mean(len_mm),
    sd_length = sd(len_mm),
    n = n(),
    se_length = sd_length / sqrt(n)
  )

print(group_summary)

# A tibble: 2 × 5
  wind  mean_length sd_length     n se_length
  <chr>       <dbl>     <dbl> <int>     <dbl>
1 lee          20.4      2.45    24     0.500
2 wind         14.9      1.91    24     0.390

Alternative 1

# Calculate the difference in means
# YOUR CODE HERE

# Assuming your dataframe is called df
group_summary %>%
  summarize(difference = mean_length[wind == "wind"] - mean_length[wind == "lee"])

# A tibble: 1 × 1
  difference
       <dbl>
1       -5.5

Alternative 2

# Or alternatively using filter and pull:
lee_mean <- group_summary %>% filter(wind == "lee") %>% pull(mean_length)
wind_mean <- group_summary %>% filter(wind == "wind") %>% pull(mean_length)
difference <- wind_mean - lee_mean
difference

[1] -5.5

Part 2: Visualizing Group Differences

Activity: Create visualizations to compare the groups Effective visualizations for group comparisons:

Side-by-side boxplots
Violin plots
Error bar plots

# YOUR TASK: Create boxplots to compare groups
ggplot(pine_data, aes(x = wind, y = len_mm, fill = wind)) +
  geom_boxplot() +
  labs(title = "Pine Needle Length by Wind Exposure",
       x = "Wind Exposure",
       y = "Length (mm)") +
  theme_minimal()

# how can you do this by wind to see both plots

your task

# YOUR TASK: Create a plot using stat_summary to show means and standard errors
ggplot(pine_data, aes(x = wind, y = len_mm, color = wind)) +
  stat_summary(fun = mean, geom = "point") +
  stat_summary(fun.data = mean_se, geom = "errorbar", width = 0.2) +
  labs(title = "Mean Pine Needle Length by Wind Exposure",
       x = "Wind Exposure",
       y = "Mean Length (mm)") +
  theme_minimal()

Part 2: Testing Assumptions for Two-Sample T-Test

Activity: Test assumptions for two-sample t-test

For a two-sample t-test, we need to check:

Normality within each group
Equal variances between groups (for standard t-test)
Independent observations

If assumptions are violated:

Welch’s t-test (unequal variances)
Non-parametric alternatives (Mann-Whitney U test)

your task

# YOUR TASK: Test normality of windward pine needle lengths
# QQ Plot
qqPlot(pine_data$len_mm, 
       main = "QQ Plot for Windward Pine Needles",
       ylab = "Sample Quantiles")

[1]  4 28

# Testing normality for each group
# Leeward group
lee_data <- pine_data %>% filter(wind == "lee")
shapiro_lee <- shapiro.test(lee_data$len_mm)
print("Shapiro-Wilk test for leeward data:")

[1] "Shapiro-Wilk test for leeward data:"

print(shapiro_lee)


    Shapiro-Wilk normality test

data:  lee_data$len_mm
W = 0.95477, p-value = 0.3425

windward group

# Windward group
# YOUR CODE HERE for windward group normality test

Remember you can always do it in one go

# there are always two ways
# Test for normality using Shapiro-Wilk test for each wind group
# All in one pipeline using tidyverse approach
normality_results <- pine_data %>%
  group_by(wind) %>%
  summarize(
    shapiro_stat = shapiro.test(len_mm)$statistic,
    shapiro_p_value = shapiro.test(len_mm)$p.value,
    normal_distribution = if_else(shapiro_p_value > 0.05, "Normal", "Non-normal")
  )

# Print the results
print(normality_results)

# A tibble: 2 × 4
  wind  shapiro_stat shapiro_p_value normal_distribution
  <chr>        <dbl>           <dbl> <chr>              
1 lee          0.955           0.343 Normal             
2 wind         0.961           0.451 Normal

Conduct a Levenes Test

# Test for equal variances
# YOUR TASK: Conduct Levene's test for equality of variances
levene_test <- leveneTest(len_mm ~ wind, data = pine_data)
print(levene_test)

Levene's Test for Homogeneity of Variance (center = median)
      Df F value Pr(>F)
group  1  1.2004 0.2789
      46

# Visual check for normality with QQ plots
# YOUR CODE HERE

Part 2: Conducting the Two-Sample T-Test

Activity: Conduct a two-sample t-test

Now we can compare the mean pine needle lengths between windward and leeward sides.

H₀: μ₁ = μ₂ (The mean needle lengths are equal) H₁: μ₁ ≠ μ₂ (The mean needle lengths are different)

Deciding between:

Standard t-test (equal variances)
Welch’s t-test (unequal variances)

Based on our Levene’s test result.

# YOUR TASK: Conduct a two-sample t-test
# Use var.equal=TRUE for standard t-test or var.equal=FALSE for Welch's t-test

# Standard t-test (if variances are equal)
t_test_result <- t.test(len_mm ~ wind, data = pine_data, var.equal = TRUE)
print("Standard two-sample t-test:")

[1] "Standard two-sample t-test:"

print(t_test_result)


    Two Sample t-test

data:  len_mm by wind
t = 8.6792, df = 46, p-value = 3.01e-11
alternative hypothesis: true difference in means between group lee and group wind is not equal to 0
95 percent confidence interval:
 4.224437 6.775563
sample estimates:
 mean in group lee mean in group wind 
          20.41667           14.91667

# Welch's t-test (if variances are unequal)
# YOUR CODE HERE

# Calculate t-statistic manually (optional)
# YOUR CODE HERE: t = (mean1 - mean2) / sqrt((s1^2/n1) + (s2^2/n2))

Part 2: Interpreting and Reporting Two-Sample T-Test Results

Activity: Interpret the results of the two-sample t-test

What can we conclude about the needle lengths on windward vs. leeward sides?

How to report this result in a scientific paper:

“A two-tailed, two-sample t-test at α=0.05 showed [a significant/no significant] difference in needle length between windward (M = …, SD = …) and leeward (M = …, SD = …) sides of pine trees, t(…) = …, p = ….”

Part 3: Paired T-Test (Extended Activity)

If we collected data in pairs (same tree, different sides), we would use a paired t-test. How would the analysis differ?

We’d calculate the difference for each pair
Test if the mean difference equals zero
The paired approach often has more statistical power

Paired t-test formula:

\(t = \frac{\bar{d}}{s_d/\sqrt{n}}\)

where:

\(\bar{d}\) is the mean difference
\(s_d\) is the standard deviation of differences
\(n\) is the number of pairs

Final Activity: Assumptions of Parametric Tests

Common assumptions for t-tests:

Normality: Data comes from normally distributed populations
Equal variances (for two-sample tests)
Independence: Observations are independent
No outliers: Extreme values can influence results

What can we do if our data violates these assumptions?

Alternatives when assumptions are violated:

Data transformation (log, square root, etc.)
Non-parametric tests
Bootstrapping approaches

Summary and Conclusions

In this activity, we’ve:

Formulated hypotheses about pine needle length
Tested assumptions for parametric tests
Conducted one-sample and two-sample t-tests
Visualized data using appropriate methods
Learned how to interpret and report t-test results

Key takeaways:

Always check assumptions before conducting tests
Visualize your data to understand patterns
Report results comprehensively
Consider alternatives when assumptions are violated

Lecture 5: Understanding P-values

A p-value is the probability of observing the sample result (or something more extreme) if the null hypothesis is true.

Common interpretations: - p < 0.05: Strong evidence against H₀ - 0.05 ≤ p < 0.10: Moderate evidence against H₀ - p ≥ 0.10: Insufficient evidence against H₀

Common misinterpretations: - p-value is NOT the probability that H₀ is true - p-value is NOT the probability that results occurred by chance - Statistical significance ≠ practical significance

Lecture 5: Type I and Type II Errors

When making decisions based on hypothesis tests, two types of errors can occur:

Type I Error (False Positive) - Rejecting H₀ when it’s actually true - Probability = α (significance level) - “Finding an effect that isn’t real”

Type II Error (False Negative) - Failing to reject H₀ when it’s actually false - Probability = β - “Missing an effect that is real”

Statistical Power = 1 - β - Probability of correctly rejecting a false H₀ - Increases with: - Larger sample size - Larger effect size - Lower variability - Higher α level

Practice Exercise 6: Interpreting P-values and Errors

Given the following scenarios, identify whether a Type I or Type II error might have occurred:

A researcher concludes that a new fishing regulation increased grayling size, when in fact it had no effect.
A study fails to detect a real decline in grayling population due to warming water, concluding there was no effect.
Let’s calculate the power of our t-test to detect a 30 mm difference in length between lakes:

# Calculate power for detecting a 30 mm difference
# First determine parameters
lake_I3 <- grayling_df %>% filter(lake == "I3")
lake_I8 <- grayling_df %>% filter(lake == "I8") 

n1 <- nrow(lake_I3)
n2 <- nrow(lake_I8)
sd_pooled <- sqrt((var(lake_I3$total_length_mm) * (n1-1) + 
                  var(lake_I8$total_length_mm) * (n2-1)) / 
                  (n1 + n2 - 2))

# Calculate power
effect_size <- 30 / sd_pooled  # Cohen's d
df <- n1 + n2 - 2
alpha <- 0.05
power <- power.t.test(n = min(n1, n2), 
                     delta = effect_size,
                     sd = 1,  # Using standardized effect size
                     sig.level = alpha,
                     type = "two.sample",
                     alternative = "two.sided")

# Display results
power


     Two-sample t test power calculation 

              n = 66
          delta = 0.6741298
             sd = 1
      sig.level = 0.05
          power = 0.9702076
    alternative = two.sided

NOTE: n is number in *each* group

Lecture 5: Summary

Key concepts covered:

Probability distributions model random phenomena
- Normal distribution is especially important
- Z-scores standardize measurements
Standard error measures precision of estimates
- Decreases with larger sample sizes
- Used to construct confidence intervals
Confidence intervals express uncertainty
- Provide plausible range for parameters
- 95% CI: mean ± 1.96 × SE
Hypothesis testing evaluates claims
- Null vs. alternative hypotheses
- P-values quantify evidence against H₀
- Consider both statistical and practical significance

Final Exercise: Comprehensive Analysis

Now that we’ve covered the key concepts, let’s perform a complete analysis of the Arctic grayling data:

# Comprehensive analysis of Arctic grayling data
# 1. Data visualization
length_boxplot <- grayling_df %>%
  ggplot(aes(x = lake, y = total_length_mm, fill = lake)) +
  geom_boxplot() +
  labs(title = "Fish Length by Lake",
       x = "Lake",
       y = "Length (mm)") +
  theme_minimal()

# 2. Compare means with t-test
length_ttest <- t.test(total_length_mm ~ lake, data = grayling_df)

# 3. Length-mass relationship
length_mass_model <- lm(mass_g ~ total_length_mm * lake, data = grayling_df)
model_summary <- summary(length_mass_model)

# 4. Display results
length_boxplot

length_ttest


    Welch Two Sample t-test

data:  total_length_mm by lake
t = -15.532, df = 161.63, p-value < 2.2e-16
alternative hypothesis: true difference in means between group I3 and group I8 is not equal to 0
95 percent confidence interval:
 -109.32342  -84.66053
sample estimates:
mean in group I3 mean in group I8 
        265.6061         362.5980

model_summary


Call:
lm(formula = mass_g ~ total_length_mm * lake, data = grayling_df)

Residuals:
     Min       1Q   Median       3Q      Max 
-151.223  -14.839   -0.764   10.670  153.130 

Coefficients:
                        Estimate Std. Error t value Pr(>|t|)    
(Intercept)            -219.3313    47.9087  -4.578 9.30e-06 ***
total_length_mm           1.3924     0.1794   7.763 8.88e-13 ***
lakeI8                 -522.5506    56.5882  -9.234  < 2e-16 ***
total_length_mm:lakeI8    1.9738     0.1972  10.009  < 2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 40.93 on 162 degrees of freedom
  (2 observations deleted due to missingness)
Multiple R-squared:  0.9644,    Adjusted R-squared:  0.9637 
F-statistic:  1461 on 3 and 162 DF,  p-value: < 2.2e-16

# 5. Calculate 95% confidence intervals for each lake
ci_results <- grayling_df %>%
  group_by(lake) %>%
  summarize(
    mean_length = mean(total_length_mm, na.rm = TRUE),
    sd_length = sd(total_length_mm, na.rm = TRUE),
    n = sum(!is.na(total_length_mm)),
    se_length = sd_length / sqrt(n),
    t_crit = qt(0.975, df = n - 1),
    margin_error = t_crit * se_length,
    ci_lower = mean_length - margin_error,
    ci_upper = mean_length + margin_error,
    .groups = "drop"
  )

# Display confidence intervals
ci_results

# A tibble: 2 × 9
  lake  mean_length sd_length     n se_length t_crit margin_error ci_lower
  <chr>       <dbl>     <dbl> <int>     <dbl>  <dbl>        <dbl>    <dbl>
1 I3           266.      28.3    66      3.48   2.00         6.96     259.
2 I8           363.      52.3   102      5.18   1.98        10.3      352.
# ℹ 1 more variable: ci_upper <dbl>

# 6. Visualize regression with confidence intervals
regression_plot <- grayling_df %>%
  ggplot(aes(x = total_length_mm, y = mass_g, color = lake)) +
  geom_point(alpha = 0.7) +
  geom_smooth(method = "lm", se = TRUE) +
  labs(title = "Length-Mass Relationship by Lake",
       x = "Length (mm)",
       y = "Mass (g)") +
  theme_minimal()

regression_plot

Based on this analysis: 1. Are there significant differences in fish length between the two lakes? 2. How does the length-mass relationship differ between lakes? 3. What conclusions can you draw about Arctic grayling in these two lakes?

Lecture 5: Error Bars and Their Interpretation

Error bars are graphical representations of the variability of data that show:

The precision of a measurement
The uncertainty around an estimate
A confidence interval for a parameter

Common types of error bars: 1. Standard Error (SE): Shows precision of the mean 2. Standard Deviation (SD): Shows variability in the data 3. Confidence Interval (CI): Shows plausible range for parameter

When interpreting graphs: - Always check what the error bars represent - Non-overlapping 95% CI bars suggest statistically significant differences - Error bars help assess both statistical and practical significance

Lecture 5: Sampling and Pseudoreplication

Pseudoreplication occurs when measurements that are not independent are analyzed as if they were independent.

A critical consideration in experimental design
Results in underestimated standard errors and confidence intervals
Leads to inflated Type I error rates (false positives)

Examples of pseudoreplication: - Measuring the same individual multiple times - Treating multiple fish from the same tank as independent - Using multiple data points from a single site

How to avoid pseudoreplication: - Identify the true experimental unit - Use appropriate statistical techniques (e.g., mixed models) - Be clear about the level of replication

Lecture 5: Practical Applications in Fish Biology

The statistical concepts we’ve covered today are essential for fisheries biologists and ecologists:

Z-scores help identify unusual fish sizes in a population
Standard error quantifies uncertainty in growth rate estimates
Confidence intervals provide plausible ranges for population parameters
Hypothesis testing evaluates effects of management practices
P-values determine significance of environmental impacts

Real-world applications: - Assessing population health and structure - Evaluating effectiveness of fishing regulations - Quantifying relationships between fish size and habitat variables - Predicting impacts of climate change on fish populations - Designing effective conservation strategies

Lecture 5: Statistical hypothesis testing

Major goal of statistics:

inferences about populations from samples assign degree of confidence to inferences

Statistical H-testing:

formalized approach to inference

hypotheses ask whether samples come from populations with certain properties
often interested in questions about population means (but not only)

Lecture 5: Statistical hypothesis testing

Relies on specifying null hypothesis (Ho) and alternate hypothesis (Ha)

Ho is the hypothesis of “no effect”
- (two samples from population with same mean, sample is from population of mean=0)
Ha (research hypothesis) the opposite of the Ho

Practice Exercise 10: Formulating Hypotheses

For the following scenarios, write out the null and alternative hypotheses:

Testing if the mean fish length in Lake S 06 is greater than 50 mm.
Testing if there is a difference in mean fish lengths between lakes Toolik and S 06.
Testing if lake E 01 has a higher variance in fish lengths compared to Lake Toolik.

For each scenario, remember that the null hypothesis typically represents “no effect” or “no difference”, while the alternative hypothesis represents what you are trying to demonstrate.

Lecture 5: Statistical hypothesis testing

p = 0.3 means that if study repeated 100 times
- would get this (or more extreme) result due to chance 30 times
p = 0.03 means that if study repeated 100 times
- would get this (or more extreme) result due to chance 3 times

Which p-value suggests Ho likely false?

Lecture 5: Statistical hypothesis testing

At what point reject Ho?

p < 0.05 conventional “significance threshold” (α)
p < 0.05 means:
- if Ho is true - if study repeated 100 times
  - would get this (or more extreme) result less than 5 times due to chance

Lecture 5: Statistical hypothesis testing

α is the rate at which we will reject a true null hypothesis (Type I error rate)
Lowering α will lower likelihood of incorrectly rejecting a true null hypothesis (e.g., 0.01, 0.001)
Both hypotheses and α are specified BEFORE collection of data and analysis

Lecture 5: Statistical hypothesis testing

Traditionally α=0.05 is used as a cut off for rejecting null hypothesis

Nothing magical about 0.0 - actual p-values need to be reported.

p-value range	Interpretation
P > 0.10	No evidence against Ho - data appear consistent with Ho
0.05 < P < 0.10	Weak evidence against the Ho in favor of Ha
0.01 < P < 0.05	Moderate evidence against Ho in favor of Ha
0.001 < P < 0.01	Strong evidence against Ho in favor of Ha
P < 0.001	Very strong evidence against Ho in favor of Ha

Lecture 5: Statistical hypothesis testing

Fisher:

p-value as informal measure of discrepancy betwen data and Ho

Lecture 5: Statistical hypothesis testing

General procedure for H testing:

Specify Null (Ho) and alternate (Ha)
Determine test (and test statistic) to be used
Test statistic is used to compare your data to expectation under Ho (null hypothesis)
Specify significance (α or p value) level below which Ho will be rejected

Lecture 5: Statistical hypothesis testing

General procedure for H testing:

Collect data - Perform test
If p-value < α, conclude Ho is likely false and reject it
If p-value > α, conclude no evidence Ho is false and retain it

Tip

Lecture 5: Next Steps in Statistical Analysis

In future lectures, we’ll explore:

One-sample and two-sample t-tests
Analysis of variance (ANOVA)
Linear regression and correlation
Chi-square tests
Non-parametric methods
Multiple regression and model selection
Mixed effects models

Each method builds on the statistical foundation we’ve established today, applying probability concepts to make inferences from data.

Learning Resources

Practice problems in the textbook (Chapter 4 & 5)
Online resources:
- Khan Academy: Probability and Statistics
- StatQuest with Josh Starmer (YouTube channel)
- R for Data Science (r4ds.had.co.nz)
Office hours: Wednesdays 2-4pm